Chapter 3: Current Electricity
Key Formulas:
• Ohm's Law: V = IR
• Resistance: R = ρL/A
• Power: P = VI = I²R = V²/R
• Kirchhoff: KVL (ΣV = 0 at junction), KCL (ΣI = 0 at junction)
• Series: R = R₁ + R₂, Parallel: 1/R = 1/R₁ + 1/R₂
Q1. A wire of resistance 12 ohms is bent in the form of a circle. Find the effective resistance between two points on any part of the circumference.
Solution
The two arcs are in parallel.
Each arc has resistance = 12/2 = 6 Ω (assuming equal halves)
1/Req = 1/6 + 1/6 = 1/3
Req = 3 Ω
Q2. A battery of emf 10V and internal resistance 3Ω is connected to a resistor. If the current in the circuit is 0.5A, find (i) the resistance of the resistor (ii) terminal voltage of the battery.
Solution
(i) I = ε/(R + r)
0.5 = 10/(R + 3)
R + 3 = 20 ⇒ R = 17 Ω
(ii) V = ε − Ir = 10 − 0.5(3) = 8.5V
Q3. A toaster of 800W and a water heater of 2000W are designed to work on 230V supply. (a) Which has greater resistance? (b) What is the resistance of each?
Solution
R = V²/P
(a) Toaster: R = 230²/800 = 52900/800 = 66.125 Ω
Heater: R = 230²/2000 = 52900/2000 = 26.45 Ω
(b) The toaster has greater resistance (lower power ⇒ higher resistance).
Q4. In the given circuit, find the current through each resistor using Kirchhoff's laws.
Solution
Consider a circuit with two loops containing batteries ε₁ = 10V, ε₂ = 5V and resistors R₁ = 2Ω, R₂ = 3Ω, R₃ = 5Ω.
By Kirchhoff's voltage law:
Loop 1: 10 − 2I₁ − 5(I₁ − I₂) = 0
Loop 2: −5 − 3I₂ − 5(I₂ − I₁) = 0
Simplify:
7I₁ − 5I₂ = 10 ... (i)
−5I₁ + 8I₂ = −5 ... (ii)
Solving: I₁ = 55/31 ≈ 1.77A, I₂ = 25/31 ≈ 0.81A
Current through R₃ = I₁ − I₂ = 30/31 ≈ 0.97A
Q5. Find the equivalent resistance between A and B of the network shown (3 resistors each of 6Ω in delta configuration).
Solution
For a delta of three equal resistors R connected between A, B, C:
Equivalent resistance between any two terminals = 2R/3
Req = 2(6)/3 = 4 Ω
Q6. The resistivity of a material is 1 × 10−7 Ωm and its dimensions are: length = 10 cm, cross-sectional area = 2 mm². Find its resistance.
Solution
R = ρL/A
L = 0.1 m, A = 2 × 10−6 m²
R = (10−7)(0.1)/(2 × 10−6)
= 10−8/(2 × 10−6)
= 0.005 Ω = 5 mΩ
Q7. In a Wheatstone bridge, P = 100 Ω, Q = 10 Ω, R = 400 Ω. Find the value of S for the bridge to be balanced.
Solution
At balance: P/Q = R/S
100/10 = 400/S
S = 400/10 = 40 Ω
Q8. A potentiometer wire of length 400 cm has a potential gradient of 0.1 V/cm. Two cells of emfs 2V and 3.5V are connected in series (opposing). Find the balance point for the combination.
Solution
When connected opposing: Vnet = 3.5 − 2 = 1.5 V
l = V/gradient = 1.5/0.1 = 15 cm
Q9. A wire of length L and resistance R is stretched to twice its original length. What will be (i) its new resistance (ii) its new resistivity?
Solution
Volume remains constant: A × 2L = A' × L ⇒ A' = A/2
(i) R' = ρ(2L)/(A/2) = 4(ρL/A) = 4R
(ii) Resistivity is a material property — it does not change. ρ' = ρ
Q10. In a meter bridge, the balance point is found at 40 cm from one end. If a 10 Ω resistor is connected in the left gap, find the resistance in the right gap.
Solution
For meter bridge: R₁/R₂ = l₁/l₂
10/R₂ = 40/60
R₂ = 10 × 60/40 = 15 Ω
Q11. A battery of emf 12V and internal resistance 2Ω is connected to a combination: R₁ = 3Ω in series with a parallel combination of R₂ = 4Ω and R₃ = 12Ω. Find (a) total current (b) terminal voltage.
Solution
Parallel: Rp = (4 × 12)/(4 + 12) = 48/16 = 3Ω
Total external R = 3 + 3 = 6Ω
(a) I = ε/(R + r) = 12/(6 + 2) = 1.5 A
(b) V = ε − Ir = 12 − 1.5(2) = 9V
Q12. What is the drift velocity of free electrons in a copper wire of cross-sectional area 1 mm² carrying a current of 2A? (Given: n = 8.5 × 1028 m−3, e = 1.6 × 10−19 C)
Solution
I = neAvd
vd = I/(neA)
= 2/((8.5 × 1028)(1.6 × 10−19)(1 × 10−6))
= 2/(1.36 × 104)
= 1.47 × 10−4 m/s
Q13. The resistance of a tungsten filament at 20°C is 100 Ω and at 1500°C is 500 Ω. Calculate the average temperature coefficient of resistance of tungsten in this range.
Solution
R = R₀(1 + αΔT)
500 = 100[1 + α(1500 − 20)]
5 = 1 + α(1480)
α = 4/1480 = 2.70 × 10−3 /°C
Q14. Two cells of emfs 1.5V and 2.5V and internal resistances 1Ω and 2Ω are connected in parallel with their positive terminals together and negative terminals together. Find the equivalent emf and internal resistance of the combination.
Solution
εeq = (ε₁/r₁ + ε₂/r₂)/(1/r₁ + 1/r₂)
= (1.5/1 + 2.5/2)/(1/1 + 1/2)
= (1.5 + 1.25)/(1.5)
= 2.75/1.5 = 1.83 V
req = 1/(1/r₁ + 1/r₂) = 1/(1 + 0.5) = 0.67 Ω